Problem: $ \left(\dfrac{1}{9}\right)^{-\frac{3}{2}}$
$= 9^{\frac{3}{2}}$ $= \left(9^{\frac{1}{2}}\right)^{3}$ To simplify $9^{\frac{1}{2}}$ , figure out what goes in the blank: $\left(? \right)^{2}=9$ To simplify $9^{\frac{1}{2}}$ , figure out what goes in the blank: $\left({3}\right)^{2}=9$ so $ 9^{\frac{1}{2}}=3$ So $9^{\frac{3}{2}}=\left(9^{\frac{1}{2}}\right)^{3}=3^{3}$ $= 3^{3}$ $= 3\cdot3\cdot 3$ $= 9\cdot3$ $= 27$